Compute how many n-digit numbers

Homework Statement

Compute how many n-digit numbers can be made from the digits of at least one of {0,1,2,3,4,5,6,7,8,9 }
Assume, repetition or order do not matter.

Homework Equations

## a_{1}, a_{2}, ..., a_{n} ##

The Attempt at a Solution

10 choices for the 1st sub-index, 10 choices for the second sub-index, ..., 10 choices for the nth- sub-index.
## 10^{n} ## total possible combinations.
I think that now we need to add 'n' for a set full of one identical digit. i.e.: {2,2,...,n-th}
Now, n*9 for all possibilities

Now, I need to take some n_i number into account in pairs and each of the n-2 numbers repeated for each number.
So, groups of two repeated for each i-th number?
Also, then I would extend this to include triple identical numbers and the rest (n-3) numbers in the set?
And, so on...?

I am sorry, if this does not make any sense or it is too messy.
Could someone give me any guidance?
Thank you.

Homework Statement

Compute how many n-digit numbers can be made from the digits of at least one of {0,1,2,3,4,5,6,7,8,9 }
Assume, repetition or order do not matter.

How can order not matter for numbers?

Homework Equations

## a_{1}, a_{2}, ..., a_{n} ##

The Attempt at a Solution

10 choices for the 1st sub-index,

Is it OK to have a string of 0's for leading digits?

10 choices for the second sub-index, ..., 10 choices for the nth- sub-index.
## 10^{n} ## total possible combinations.
I think that now we need to add 'n' for a set full of one identical digit. i.e.: {2,2,...,n-th}
Now, n*9 for all possibilities

Now, I need to take some n_i number into account in pairs and each of the n-2 numbers repeated for each number.
So, groups of two repeated for each i-th number?
Also, then I would extend this to include triple identical numbers and the rest (n-3) numbers in the set?
And, so on...?

I don't follow what you are doing in this last section. Is there something about the problem you haven't told us?

You might be right, but he said that it does not matter. My professor was very vague in posing the question. I asked it twice.
He gave some examples:
## {0, 0, 0,0, ...,0_{n} }={0} ##
## {0, 0, 0, ..., 0, 1_{n}}={0,1}##

It turns out that order does matter after all. One 'gotta' love language.
An example helped to elucidate.

Example:
{0,1,2,3,...,8,9, ...<whatever>,...} is a vector of size n and 1 solution that satisfies the constraints.
{3,4,7,...,9,2,1,...<whatever>,... } is another vector of size n and another solution.

Let me work it out. If you have any leads, they are welcome.
Thanks :>

Please, check the solution in attachment.
Apparently, it is incorrect. Can someone verify?
I think that I am not taking into account cases such as {m, o ,m } or {1,0,1}, where there could be repetitions.

The solution should be in the form:
Order matters
{ ... , <at least digits from 0 to 9>,..., <any numbers>, ... }

Thank you.

Attachments

## \dfrac{ |n| !}{ |n_{1}|! |n_{2}|!... |n_{k}|!} \text{, where n is the size of the vector and the values in denominator are types of symbols in n that repeat.}## ##\text{For instance, if I have a vector called v={e, i, g, e, n, v, a, l, u, e}, there are say n1 type symbols.}## ##\text{n1 relates to, say, e and our |n1|=3. The numerator in the equation takes care of all combination including repetitions}## ##\text{and the denominator takes care of cases as the one just mentioned. } ##